# Euler-Lagrange equation problem

For the following variational problem I have been told to show the Euler-Lagrange equation is identically zero.

$$L[u]:= \int_a^b \sin(u)u_x\,\mathrm dx$$

I have found it to be

$u_x\cos(u)-\sin(u)u_{xx}=0.$

Is this correct? And if so, does this always equal $0$?

• I think your second term is wrong, it should be something like $\frac{d}{dx} \sin(u)$, which will end up canceling the first term. – asperanz Oct 21 '12 at 20:01

## 2 Answers

No: write $F(t_1,t_2,t_3):=t_3\sin t_2$. Then $L(u)=\int_a^bf(t,u(t),u'(t))dt.$ Euler-Lagrange equation is $\partial_{2}F(t,u,u')=\frac d{dt}\partial_3F(t,u,u')$, hence $$u'(t)\cos u(t)-\frac d{dx}\sin(u(t))=0,$$ which is always satisfied.

Your equation of motion is not correct, you were already given the right one. But indeed your equation is trivially satisfied. That's because the Lagrangian itself is a total derivative, then the action

$$L[q]=-\int_a^b\frac{d}{dt}\cos(q(t))\mathrm{d}t=cte.$$

does not depend on the path $q(t)$ but on the endpoints, which you held fixed. That is, as a variational problem, it is trivial as any Lagrangian having the form $\mathcal{L}[q,t]=\dot{f}(q(t))$. Actually, Lagrangians are not uniquelly, but defined up to those kind of terms.

In higher dimensions the Lagrangians that mimic that in your question are

$$\mathcal{L}[\phi]=\mathrm{d}{\phi}$$

for $\phi$ a (dim$M-1$)-form, so that, by Stokes theorem:

$$S=\int_M\mathcal{L} =\int_M\mathrm{d}\phi = \int_{\partial M} \phi$$

If $M$ is boundaryless or $\phi$ is compactly supported, this term has trivial equations of motion $-$ no dynamics.